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Electricity and Magnetism
Mathematical Tools and Fields
01Series Approximations: Keeping One or Two Terms02Integrals as Sums: Along a Line and a Curved Path03Line Integrals: Work as the Integral of F · ds04Closed Line Integrals: Work Around a Loop05Double Integrals: Area, Charge and Volume06Closed Surface Integrals: The Sphere and Its Charge07Electric Flux Through Surfaces and Volume IntegralsProblem set0/10Problem set 20/10MIT problem set0/4
01Field Theory Versus Action at a Distance02Scalar Fields: Contour, Color and Relief Maps03Vector Fields, Sources, Sinks and Divergence04Curl: Measuring the Circulation of a Field05Gravitational, Electric and Magnetic Fields06Drawing Vector Fields: Arrows, Lines and Grass SeedsProblem set0/10Problem set 20/10MIT problem set0/1
Electrostatics
01Coulomb's Law: Force Between Two Point Charges02Superposition: Adding Coulomb Forces as Vectors03Field Lines, Null Points and Uniform Fields04The Electric Dipole and Its 1/r³ Field05Torque, Energy and Force on a Dipole in a FieldProblem set0/10Problem set 20/10MIT problem set0/1
01Charge Density and the Field on a Rod's Axis02The Field on a Charged Rod's Perpendicular Bisector03The Fields of a Charged Ring and a Charged Disk04How Field Lines Carry Force, Energy and Momentum05Electric Pressure and Tension: The Stress ε₀E²/2Problem set0/10Problem set 20/10MIT problem set0/6

Charge Density and the Field on a Rod's Axis

How do charge densities turn a rod's field into an integral, and when does the rod look like a point charge?


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Your summary note

    1. 1

      Volume, surface and line charge densities ρ\rhoρ, σ\sigmaσ, λ\lambdaλ

      Define each density as ρ=dqdV\rho = \frac{dq}{dV}ρ=dVdq​, σ=dqdA\sigma = \frac{dq}{dA}σ=dAdq​, λ=dqdℓ\lambda = \frac{dq}{d\ell}λ=dℓdq​ with its SI unit, and write the integral of each density that gives the total charge QQQ.

    2. 2

      The field of a continuous distribution as a vector integral

      Write Coulomb's law for one element dqdqdq, sum the contributions into E⃗=14πε0∫dqr2r^\vec{E} = \frac{1}{4\pi\varepsilon_0}\int \frac{dq}{r^2}\hat{r}E=4πε0​1​∫r2dq​r^, and record that this stands for three separate component integrals.

    3. 3

      Worked example: the field on a uniform rod's axis

      Set up the rod from x0x_0x0​ to x0+ℓx_0+\ellx0​+ℓ with dq=λ dx′dq = \lambda\,dx'dq=λdx′ and r^=−i^\hat{r} = -\hat{i}r^=−i^, integrate to E⃗=−Q4πε0x0(ℓ+x0)i^\vec{E} = -\frac{Q}{4\pi\varepsilon_0 x_0(\ell+x_0)}\hat{i}E=−4πε0​x0​(ℓ+x0​)Q​i^, and recover the point-charge field when x0≫ℓx_0 \gg \ellx0​≫ℓ.

    Attempt 1 of 2